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3 months ago

. Andrew made an error in determining the polynomial equation of smallest degree whose roots are 3, 2+2i

Mathematics

1 answer:

PIT_PIT [12.4K]3 months ago

7 0

Answer:

Error made by Andrew: He identified incorrect factors based on the roots.

Step-by-step explanation:

The roots of the polynomial consist of: 3, 2 + 2i, 2 - 2i. By the factor theorem, if a is a root of the polynomial P(x), then (x - a) must be a factor of P(x). According to this premise:

(x - 3), (x - (2 + 2i)), (x - (2 - 2i)) represent the factors of the polynomial.

<pBy simplification, we obtain:

(x - 3), (x - 2 - 2i), (x - 2 + 2i) as the respective factors.

This is where Andrew's mistake occurred. Factors should always be in the form (x - a), not (x + a). Andrew expressed the complex factors incorrectly, resulting in an erroneous conclusion.

Thus, the polynomial can be expressed as:

$(x - 3)(x - 2 - 2i)(x - 2+2i)=0\\\\ (x-3)(x^{2}-2x+2xi-2x+4-4i-2xi+4i-4i^{2})=0\\\\ (x-3)(x^{2}-4x+4+4)=0\\\\ (x-3)(x^{2}-4x+8)=0\\\\ x^{3}-4x^{2}+8x-3x^{2}+12x-24=0\\\\ x^{3}-7x^{2}+20x-24=0$

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